Problem 124 - A Graphical Approach to Precalculus with Limits

Question

Solve each formula for the indicated variable. Leave \(\pm\) in answers when applicable. Assume that no denominators are 0 $$V=\frac{4}{3} \pi r^{3} \quad \text { for } r$$

Step-by-Step Solution

Answer

\( r = \pm \sqrt[3]{\frac{3V}{4\pi}} \)

Step 1: Identify the given formula

The formula given in the problem is the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). We are asked to solve for \( r \), the radius.

Step 2: Isolate the term with the variable

To solve for \( r \), we need to isolate \( r^3 \). Start by multiplying both sides of the equation by \( \frac{3}{4\pi} \) to cancel \( \frac{4}{3}\pi \) on the right side. This gives: \( r^3 = \frac{3V}{4\pi} \).

Step 3: Solve for the variable

To find \( r \), take the cube root of both sides of the equation from Step 2. That results in: \( r = \sqrt[3]{\frac{3V}{4\pi}} \).

Step 4: Include the \(\pm\) sign

Since the cube root can result in both positive and negative solutions (in real-world contexts only positive is physically meaningful but mathematically both are solutions), express \( r \) as \( r = \pm\sqrt[3]{\frac{3V}{4\pi}} \).

Key Concepts

Sphere Volume Formula

The formula for the volume of a sphere is an essential tool in geometry and precalculus. It tells us how much space is inside a sphere. The formula is given as: \[ V = \frac{4}{3} \pi r^3 \]where:

• \( V \) is the volume of the sphere
• \( \pi \) is a constant approximately equal to 3.14159
• \( r \) is the radius of the sphere

This formula comes from mathematical derivations involving the properties of circles and spheres. Understanding it gives insight into how dimensions of a sphere relate to the space it occupies.

Isolating Variables

In algebra, isolating variables is a technique used to solve equations for a given variable. It's like solving a mystery, focusing on one clue at a time. When you have an equation, the goal is to get the variable of interest on one side alone.
For the formula of the sphere's volume, we isolated \( r^3 \). We multiplied both sides by \( \frac{3}{4\pi} \) to cancel out the fractional coefficient of \( r^3 \).
This gives us:\[ r^3 = \frac{3V}{4\pi} \]Thus, isolating the variable is a crucial step that makes it possible to apply additional mathematical operations to find the exact value of \( r \).

Cube Root

Taking the cube root is a method to undo cubed numbers, similar to how a square root undoes squaring a number. When you have a number like \( r^3 \), taking the cube root (\( \sqrt[3]{\cdot} \)) brings it back to \( r \).
In our exercise, once we isolated \( r^3 \), we took the cube root of both sides:\[ r = \sqrt[3]{\frac{3V}{4\pi}} \]This operation simplifies the equation and helps us directly find \( r \), demonstrating how cube roots can solve problems involving cubic terms.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations using mathematical operations. It's essential for deriving solutions in many mathematical contexts.
Throughout solving the sphere volume formula for \( r \), we employed algebraic manipulation:

• Multiplying both sides by \( \frac{3}{4\pi} \) to isolate \( r^3 \)
• Taking the cube root of both sides to solve for \( r \)
• Introducing the \( \pm \) symbol to indicate potential positive and negative results

Understanding these steps helps in performing similar manipulations in other equations, showcasing the flexibility and utility of algebra.